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\@writefile{toc}{\contentsline {subsection}{\numberline {2.2}函数 $f(x)=\left \{\begin  {aligned}&0 & x=0 \\ &\frac  {1}{n} & \frac  {1}{n+1}< x \leqslant \frac  {1}{n}, n=1,2, \cdots  \end  {aligned}\right .$,在区间$[0,1]$上可积}{1}\protected@file@percent }
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\@writefile{toc}{\contentsline {section}{\numberline {3}讨论以下命题}{2}\protected@file@percent }
\@writefile{toc}{\contentsline {subsection}{\numberline {3.1}设函数 $f(x)$ 在 $[a, b]$ 上可积,且满足 $|f(x)| \geqslant m>0$$m $ 为一常数), 证明函数 $\frac  {1}{f(x)}$ 在 $[a, b]$ 上可积}{2}\protected@file@percent }
\@writefile{toc}{\contentsline {subsection}{\numberline {3.2}设函数 $f(x), g(x)$ 在 $[a, b]$ 上可积,复合函数 $f(g(x))$ 在 $[a, b]$ 上是否一定可积? 当函数 $f(x)$ 为连续函数是结论是否成立}{2}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {4}证明:若 $f(x)$ 在 $[a, b]$ 上连续,且 $f(x) \geqslant 0,$ 若 $\DOTSI \intop \ilimits@ _{a}^{b} f(x) \mathrm  {\nobreakspace  {}d} x=0 .$ 则 $f(x) \equiv 0, x \in [a, b]$}{2}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {5}比较下列积分的大小:}{3}\protected@file@percent }
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\@writefile{toc}{\contentsline {section}{\numberline {6}设 $f(x)$ 在 $[a, b]$ 上可积,则任给 $\varepsilon >0,$ 存在阶梯函数 $g(x),$ 使得$\DOTSI \intop \ilimits@ _{a}^{b}|f(x)-g(x)| \mathrm  {\nobreakspace  {}d} x<\varepsilon $.证明 : 由于 $f(x) \in R[a, b],$ 对于 $\forall \varepsilon >0,$ 存在 $[a, b]$ 的分割$\Delta : a=x_{0}<x_{1}<\cdots  <x_{n}=b$}{3}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {7}求下列极限:}{5}\protected@file@percent }
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\@writefile{toc}{\contentsline {subsection}{\numberline {8.1}$\frac  {\pi }{2}<\DOTSI \intop \ilimits@ _{0}^{\pi / 2} \frac  {\mathrm  {\nobreakspace  {}d} x}{\sqrt  {1-\frac  {1}{2} \qopname  \relax o{sin}^{2} x}}<\frac  {\pi }{\sqrt  {2}}$}{5}\protected@file@percent }
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\@writefile{toc}{\contentsline {section}{\numberline {9}设 $f(x)$ 在 [0,1] 上连续,且单调递减,证明对任意$\beta \in [0,1],$ 都有$\DOTSI \intop \ilimits@ _{0}^{\beta } f(x) \mathrm  {\nobreakspace  {}d} x \geqslant \beta \DOTSI \intop \ilimits@ _{0}^{1} f(x) \mathrm  {\nobreakspace  {}d} x$}{6}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {10}设函数 $f(x), g(x)$ 在 $[a, b]$ 上可积,且 $f(x), g(x)$ 仅在有限个点处不相等,则有$\DOTSI \intop \ilimits@ _{a}^{b} f(x) \mathrm  {d} x=\DOTSI \intop \ilimits@ _{a}^{b} g(x) \mathrm  {d} x$}{6}\protected@file@percent }
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\@writefile{toc}{\contentsline {subsection}{\numberline {11.1}设 $f(x)$ 在 $[a, b]$ 上可积,则$$\left (\DOTSI \intop \ilimits@ _{a}^{b} f(x) \qopname  \relax o{sin}x \mathrm  {\nobreakspace  {}d} x\right )^{2}+\left (\DOTSI \intop \ilimits@ _{a}^{b} f(x) \qopname  \relax o{cos}x \mathrm  {\nobreakspace  {}d} x\right )^{2} \leqslant (b-a) \DOTSI \intop \ilimits@ _{a}^{b} f^{2}(x) \mathrm  {d} x$$}{7}\protected@file@percent }
\@writefile{toc}{\contentsline {subsection}{\numberline {11.2}设 $f(x)$ 在 $[a, b]$ 上可积且非负,则$$\left (\DOTSI \intop \ilimits@ _{a}^{b} f(x) \qopname  \relax o{sin}n x \mathrm  {\nobreakspace  {}d} x\right )^{2}+\left (\DOTSI \intop \ilimits@ _{a}^{b} f(x) \qopname  \relax o{cos}n x \mathrm  {\nobreakspace  {}d} x\right )^{2} \leqslant \left (\DOTSI \intop \ilimits@ _{a}^{b} f(x) \mathrm  {d} x\right )^{2}$$}{7}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {12}引理1——$Young$不等式:$\frac  {1}{p}+\frac  {1}{q}=1$,则有$x^{\frac  {1}{p}} y^{\frac  {1}{q}} \leqslant \frac  {x}{p}+\frac  {y}{q}$}{7}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {13}引理2——$\mathrm  {Hölder}$不等式:设 $a_{i}>0, b_{i}>0(i=1,2, \cdots  , n),$ 则$$\DOTSB \sum@ \slimits@ _{i=1}^{n} a_{i} b_{i} \leqslant \left (\DOTSB \sum@ \slimits@ _{i=1}^{n} a_{i}^{p}\right )^{\frac  {1}{p}}\left (\DOTSB \sum@ \slimits@ _{i=1}^{n} b_{i}^{q}\right )^{\frac  {1}{q}}$$}{7}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {14}引理3——$\mathrm  {Minkowskii}$不等式:设 $a_{1}, a_{2}, \cdots  , a_{n}$ 和 $b_{1}, b_{2}, \cdots  , b_{n}$ 是两组非负实数, $p>1,$ 则$$\left (\DOTSB \sum@ \slimits@ _{i=1}^{n}\left (a_{i}+b_{i}\right )^{p}\right )^{\frac  {1}{p}} \leqslant \left (\DOTSB \sum@ \slimits@ _{i=1}^{n} a_{i}^{p}\right )^{\frac  {1}{p}}+\left (\DOTSB \sum@ \slimits@ _{i=1}^{n} b_{i}^{p}\right )^{\frac  {1}{p}}$$}{8}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\numberline {15}证明下列不等式}{9}\protected@file@percent }
\@writefile{toc}{\contentsline {subsection}{\numberline {15.1}赫尔德($\mathrm  {Hlder}$)不等式:设 $f, g$ 在 $[a, b]$ 上连续, $p>1, \frac  {1}{p}+\frac  {1}{q}=1,$ 则$$\DOTSI \intop \ilimits@ _{a}^{b}|f(x) g(x)| \mathrm  {d} x \leqslant \left [\DOTSI \intop \ilimits@ _{a}^{b}|f(x)|^{p} \mathrm  {\nobreakspace  {}d} x\right ]^{\frac  {1}{p}}\left [\DOTSI \intop \ilimits@ _{a}^{b}|g(x)|^{q} \mathrm  {\nobreakspace  {}d} x\right ]^{\frac  {1}{q}}$$}{9}\protected@file@percent }
\@writefile{toc}{\contentsline {subsection}{\numberline {15.2}闵科夫斯基($\mathrm  {Minkowski}$)不等式 :设 $f, g$ 在 $[a, b]$ 上连续, $p>1,$ 则$$\left (\DOTSI \intop \ilimits@ _{a}^{b}|f(x)+g(x)|^{p} \mathrm  {\nobreakspace  {}d} x\right )^{\frac  {1}{p}} \leqslant \left (\DOTSI \intop \ilimits@ _{a}^{b}|f(x)|^{p} \mathrm  {\nobreakspace  {}d} x\right )^{\frac  {1}{p}}+\left (\DOTSI \intop \ilimits@ _{a}^{b}|g(x)|^{p} \mathrm  {\nobreakspace  {}d} x\right )^{\frac  {1}{p}}$$}{10}\protected@file@percent }
